Third Edition
Janice Gorzynski Smith
University of Hawai’i
Prepared by Layne A. Morsch
The University of Illinois - Springfield
Copyright © 2011 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1

Alkene Structure
• Alkenes are also called olefins .
• Alkenes contain a carbon-carbon double bond.
• Terminal alkenes have the double bond at the end of the carbon chain.
•
Internal alkenes have at least one carbon atom bonded to each end of the double bond.
• Cycloalkenes contain a double bond in a ring.
2

Alkene Hybridization
• Recall that the double bond consists of a
 bond and a
 bond.
• Each carbon is sp 2 hybridized and trigonal planar, with bond angles of approximately 120
°.
3

Bond Dissociation Energy
• Bond dissociation energies of the C-C bonds in ethane (a
 bond only) and ethylene (one
 and one
 bond) can be used to estimate the strength of the
 component of the double bond.
• The
 bond is much weaker than the
 bond of a C-C double bond, making it much more easily broken.
• Therefore, alkenes undergo many reactions that alkanes do not.
4

Cyclic Alkenes
• Cycloalkenes having fewer than eight carbon atoms have a cis geometry.
• A trans-cycloalkene must have a carbon chain long enough to connect the ends of the double bond without introducing too much strain.
• trans-Cyclooctene is the smallest, isolable trans cycloalkene, but it is considerably less stable than cis-cyclooctene, making it one of the few alkenes having a higher energy trans isomer.
5

6

Calculating Degrees of Unsaturation
• An acyclic alkene has the general structural formula C n
H
2n
.
• Alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.
• Cycloalkanes also have the general formula C n
H
2n
.
• Each
 bond or ring removes two hydrogen atoms from a molecule, and this introduces one degree of unsaturation.
• The number of degrees of unsaturation for a given molecular formula can be calculated by comparing the actual number of H atoms in a compound to the maximum number of H atoms possible for the number of carbons present if the molecule were an acyclic alkane.
• This procedure gives the total number of rings and/or
 bonds in a molecule.
7

Degrees of Unsaturation for Molecules
Containing Heteroatoms
• Ignore O atoms in the molecule (this divalent atom is a linker and has no effect on degree of unsaturation).
• Add number of halogens to number of H's (they are equivalent to H).
• Subtract 1 H for each N present (N’s two connections allows extra H).
• E.g., C
6
H
10
OCl
3
N is equivalent to C
6
H
12
.

9

Figure 10.1
Naming Alkenes and Alkenols
• Compounds that contain both a double bond and a hydroxy group are named as alkenols and the chain (or ring) is numbered to give the OH group the lower number.
10

Naming Polyenes and Cyclic Alkenes
• Compounds with two double bonds are named as dienes by changing the “-ane” ending of the parent alkane to the suffix
“-adiene”.
• Compounds with three double bonds are named as trienes, and so forth.
• In naming cycloalkenes, the double bond is located between
C1 and C2, and the “1” is usually omitted in the name.
• The ring is numbered clockwise or counterclockwise to give the first substituent the lower number.
Figure 10.2
11

12

Common Names of Alkenes and Alkene
Substituents
• Some alkene or alkenyl substituents have common names.
• The simplest alkene, CH
2
=CH
2
, named in the IUPAC system as ethene, is often called ethylene .
Figure 10.3
13

Physical Properties of Alkenes
• Most alkenes exhibit only weak van der Waals interactions, so their physical properties are similar to alkanes of comparable molecular weight.
• Alkenes have low melting points and boiling points.
• Melting and boiling points increase as the number of carbons increases because of increased surface area.
• Alkenes are soluble in organic solvents and insoluble in water.
• The C-C single bond between an alkyl group and one of the double bond carbons of an alkene is slightly polar because the sp 3 hybridized alkyl carbon donates electron density to the sp 2 hybridized alkenyl carbon.
14

Cis/Trans Differ in Physical Properties
• A consequence of the alkene dipole is that cis and trans isomeric alkenes often have somewhat different physical properties.
• cis-2-Butene has a higher boiling point (4 °C) than trans-2-butene
(1 °C).
• In the cis isomer, the two C
sp3
C
sp2 bond dipoles reinforce each other, yielding a small net molecular dipole.
• In the trans isomer, the two bond dipoles cancel.
15

Useful Products Formed From Ethylene
Figure 10.4
16

Figure 10.5
Naturally Occurring Alkenes
17

Fatty Acids
• Triacylglycerols are hydrolyzed to glycerol and three fatty acids of general structure RCOOH.
• Saturated fatty acids have no double bonds in their long hydrocarbon chains, and unsaturated fatty acids have one or more double bonds in their hydrocarbon chains.
• As the number of double bonds in the fatty acid increases, the melting point decreases.
18

• Increasing the number of double bonds in the fatty acid side chains decreases the melting point of the triacylglycerol.
19

3-D Structure of C
18
Fatty Acids
• The larger the number of Z double bonds, the more kinks in the hydrocarbon chain.
• This causes poorer stacking and less van der Waals interactions, leading to lower melting points.
Figure 10.6
20

Triacylglycerols
• Fats and oils are both triacylglycerols, but with different physical properties.
• Fats have higher melting points —they are solids at room temperature.
• Oils have lower melting points —they are liquids at room temperature.
• The composition (saturated vs. unsaturated) of the three fatty acids in the triacylglycerol determines whether it is a fat or an oil.
21

Properties of Fatty Acids
• Fats are derived from fatty acids having few or no double bonds.
• Oils are derived from fatty acids having a larger number of double bonds.
• Saturated fats are typically obtained from animal sources, whereas unsaturated oils are common in vegetable sources.
• An exception to this generalization is coconut oil, which is largely composed of saturated alkyl side chains.
22

Preparation of Alkenes
• Alkenes can be prepared from alkyl halides, tosylates, and alcohols via elimination reactions.
23

Regioselectivity and Stereoselectivity of
Alkene Formation
• The most stable alkene (Zaitsev product) is usually formed as the major product.
24

Addition Reactions
• The characteristic reaction of alkenes is addition —the  bond is broken and two new
 bonds are formed.
• Alkenes are electron rich, with the electron density of the
 bond concentrated above and below the plane of the molecule.
• Therefore, alkenes act as nucleophiles and react with electrophiles.
• Simple alkenes do not react with nucleophiles or bases, reagents that are themselves electron rich.
25

Syn and Anti Addition to Alkenes
• Because the carbon atoms of a double bond are both trigonal planar, the elements of X and Y can be added to them from the same side or from opposite sides.
• Syn addition takes place when both X and Y are added from the same side.
• Anti addition takes place when X and Y are added from opposite sides.
26

Addition Reactions of Cyclohexene
Figure 10.8
27

Hydrohalogenation —Electrophilic Addition of HX
• Two bonds are broken in this reaction —the weak  bond of the alkene and the HX bond —and two new  bonds are formed — one to H and one to X.
•
Recall that the H-X bond is polarized, with a partial positive charge on H.
• Because the electrophilic H end of HX is attracted to the electron-rich double bond, these reactions are called electrophilic additions.
28

How to Draw the Products of an
Addition Reaction
• Locate the C-C double bond.
•
Identify the
 bond of the reagent that breaks.
• Break the
 bond of the alkene and the
 bond of the reagent, and form two new
 bonds to the C atoms of the double bond.
29

Heat of Formation for Electrophilic Addition
• Addition reactions are exothermic because the two
 bonds formed in the product are stronger than the
 and
 bonds broken in the reactants.
• For example,

H ° for the addition of HBr to ethylene is –14 kcal/mol.
Figure 10.9
30

Mechanism of Electrophilic Addition
• The mechanism of electrophilic addition consists of two successive Lewis acid-base reactions.
• Step [1] – the alkene is the Lewis base that donates an electron pair to H-Br, the Lewis acid.
• Step [2] – Br¯ is the Lewis base that donates an electron pair to the carbocation, the Lewis acid.
31

Energy Diagram for Electrophilic Addition
• Each step has its own energy barrier with a transition state energy maximum.
• Since step [1] has a higher energy transition state, it is ratedetermining.
• 
H
° for step [1] is positive because more bonds are broken than formed, whereas

H
° for step [2] is negative because only bond making occurs.
Figure 10.10
32

Markovnikov’s Rule
• With an unsymmetrical alkene, HX can add to the double bond to give two constitutional isomers, but only one is actually formed:
• Markovnikov’s rule states that in the addition of HX to an unsymmetrical alkene, the H atom adds to the less substituted carbon atom —that is, the carbon that has the greater number of H atoms to begin with.
33

Carbocation Stability and Markovnikov’s Rule
• The basis of Markovnikov’s rule is the formation of a carbocation in the rate-determining step of the mechanism.
• In the addition of HX to an unsymmetrical alkene, the H atom is added to the less substituted carbon to form the more stable, more substituted carbocation.
34

Hammond Postulate and Electrophilic
Addition
• According to the Hammond postulate, Path [2] is faster because formation of the carbocation is an endothermic process.
• Thus, the transition state to form the more stable 2 ° carbocation is lower in energy.
• The E a for formation of the more stable 2 ° carbocation is lower than the E a for formation of the 1 ° carbocation; the 2° carbocation is formed faster.
Figure 10.11
35

Stereochemistry of Electrophilic Addition
• Recall that trigonal planar atoms react with reagents from two directions with equal probability.
• Achiral starting materials yield achiral products.
• Sometimes new stereogenic centers are formed from hydrohalogenation.
36

Stereochemistry of Carbocation Formation
• The mechanism of hydrohalogenation illustrates why two enantiomers are formed.
• Initial addition of H
+ occurs from either side of the planar double bond.
• Both modes of addition generate the same achiral carbocation.
• Either representation of this carbocation can be used to draw the second step of the mechanism.
37

Stereochemistry of Nucleophilic Attack
• Nucleophilic attack of Cl ¯ on the trigonal planar carbocation also occurs from two different directions, forming two products, A and
B, having a new stereogenic center.
• A and B are enantiomers.
• Since attack from either direction occurs with equal probability, a racemic mixture of A and B is formed.
38

Reaction of 1,2-dimethylcyclohexene with HCl
• Addition of HX to 1,2-dimethylcyclohexene forms two new stereogenic centers.
• Four stereoisomers are formed:
•
Compounds A and D are enantiomers.
• Compounds B and C are enantiomers.
Figure 10.12
39

Hydrohalogenation —Summary
40

Hydration —Electrophilic Addition of Water
• Hydration is the addition of water to an alkene to form an alcohol.
41

Mechanism of Hydration
42

Electrophilic Addition of Alcohols
• Alcohols add to alkenes, forming ethers by the same mechanism.
• For example, addition of CH
3
OH to 2-methylpropene, forms
tert-butyl methyl ether (MTBE), a high octane fuel additive.
•
Note that there are three consequences to the formation of carbocation intermediates:
1.
Markovnikov’s rule holds.
2. Addition of H and OH occurs in both syn and anti fashion.
3. Carbocation rearrangements can occur.
43

Halogenation —Electrophilic Addition of Halogen
• Halogenation is the addition of X
2 alkene to form a vicinal dihalide.
(X = Cl or Br) to an
44

Halogenation Details
• Halogens add to
 bonds because halogens are polarizable.
• The electron-rich double bond induces a dipole in an approaching halogen molecule, making one halogen atom electron deficient and the other electron rich (X

+
X
 –
).
• The electrophilic halogen atom is then attracted to the nucleophilic double bond, making addition possible.
• Two facts demonstrate that halogenation follows a different mechanism from that of hydrohalogenation or hydration.
• No rearrangements occur.
• Only anti addition of X
2 is observed.
• These facts suggest that carbocations are not intermediates.
45

Halogenation Mechanism
46

Stability of Cation Intermediates
• Carbocations are unstable because they have only six electrons that surround carbon.
• Halonium ions are unstable because of ring strain.
47

Stereochemistry of Halonium Formation
• Consider the chlorination of cyclopentene to afford both enantiomers of trans-1,2-dichlorocyclopentane, with no cis products.
• Initial addition of the electrophile Cl
+ from (Cl
2
) occurs from either side of the planar double bond to form a bridged chloronium ion.
48

Stereochemistry of Halonium Ring Opening
• In the second step, nucleophilic attack of Cl ¯ must occur from the backside.
• Since the nucleophile attacks from below and the leaving group departs from above, the two Cl atoms in the product are oriented trans to each other.
• Backside attack occurs with equal probability at either carbon of the three-membered ring to yield a racemic mixture.
49

Stereochemical Outcome of Halogenation
• cis-2-Butene yields two enantiomers, whereas trans-2-butene yields a single achiral meso compound.
Figure 10.13
50

Halohydrin Formation
• Treatment of an alkene with a halogen X
2 and H
2
O forms a halohydrin by addition of the elements of X and OH to the double bond.
51

Mechanism of Halohydrin Formation
• Even though X¯ is formed in step [1] of the mechanism, its concentration is small compared to H
2 so H
2
O and not X¯ is the nucleophile.
O (often the solvent),
52

Generating Bromine in Halohydrin Formation
• Although the combination of Br
2 and H
2
O effectively forms bromohydrins from alkenes, other reagents can also be used.
• Bromohydrins are also formed with N-bromosuccinimide
(NBS) in aqueous DMSO [(CH
3
)
2
S=O].
• In H
2
O, NBS decomposes to form Br
2
, which then goes on to form a bromohydrin by the same reaction mechanism.
53

Anti Stereochemistry in Halohydrin Formation
• Because the bridged halonium ion is opened by backside attack of H
2
O, addition of X and OH occurs in an anti fashion and trans products are formed.
• With unsymmetrical alkenes, the preferred product has the electrophile X+ bonded to the less substituted carbon, and the nucleophile (H
2
O) bonded to the more substituted carbon.
54

Regiochemistry of Halohydrin Formation
• As in the acid catalyzed ring opening of epoxides, nucleophilic attack occurs at the more substituted carbon end of the bridged halonium ion because that carbon is better able to accommodate the partial positive charge in the transition state.
55

Summary of Halohydrin Formation
56

Halohydrin Use in Synthesis
• Halohydrins have been used in the synthesis of many naturally occurring compounds.
• Key steps in the synthesis of estrone, a female sex hormone, are illustrated below.
Figure 10.14
The synthesis of estrone from a chlorohydrin
57

Hydroboration –Oxidation
• Hydroboration –oxidation is a two-step reaction sequence that converts an alkene into an alcohol.
58

Borane and Hydroboration
• BH
3 is a reactive gas that exists mostly as a dimer, diborane
(B
2
H
6
).
• Borane is a strong Lewis acid that reacts readily with Lewis bases.
• For ease of handling in the laboratory, it is commonly used as a complex with tetrahydrofuran (THF).
• The first step in hydroboration –oxidation is the addition of the elements of H and BH
2 to the
 bond of the alkene, forming an intermediate alkylborane.
59

Hydroboration Mechanism
• The proposed mechanism involves concerted addition of H and BH
2 from the same side of the planar double bond: the
 bond and H-BH
2 bond are broken as two new
 bonds are formed.
• Because four atoms are involved, the transition state is said to be four-centered.
60

Reactivity of Borane During Hydroboration
• Because the alkylborane formed by the reaction with one equivalent of alkene still has two B-H bonds, it can react with two more equivalents of alkene to form a trialkylborane.
•
We often draw hydroboration as if addition stopped after one equivalent of alkene reacts with BH
3
.
• Instead all three B-H bonds actually react with three equivalents of an alkene to form a trialkylborane.
Figure 10.15
61

Other Sources of B-H for Hydroboration
• Since only one B-H bond is needed for hydroboration, commercially available dialkylboranes having the general structure R
2
BH are sometimes used instead of BH
3
.
•
A common example is 9-borabicyclo[3.3.1]nonane (9-BBN).
62

Regiochemistry of Hydroboration
• With unsymmetrical alkenes, the boron atom bonds to the less substituted carbon atom.
• This regioselectivity can be explained by considering steric factors.
• The larger boron atom bonds to the less sterically hindered, more accessible carbon atom.
63

Electronic Factors Affecting Regiochemistry of Hydroboration
• Electronic factors are also used to explain this regioselectivity.
• If bond making and bond breaking are not completely symmetrical, boron bears a

- charge in the transition state and carbon bears a

+ charge.
• Since alkyl groups stabilize a positive charge, the more stable transition state has the partial positive charge on the more substituted carbon.
64

Example of Regiochemistry of Hydroboration
Figure 10.16
65

Oxidation Following Hydroboration
• Since alkylboranes react rapidly with water and spontaneously burn when exposed to air, they are oxidized, without isolation, with basic hydrogen peroxide (H
2
O
2
, ¯OH).
•
Oxidation replaces the C-B bond with a C-O bond, forming a new OH group with retention of configuration.
• The overall result of this two-step sequence is syn addition of the elements of H and OH to a double bond in an “anti-
Markovnikov” fashion.
66

Summary of Hydroboration —Oxidation
67

Use of Alkenes in Synthesis
• Suppose we wish to synthesize 1,2-dibromocyclohexane from cyclohexanol.
To solve this problem we must:
68

Retrosynthetic Analysis
Working backwards from the product to determine the starting material from which it is made is called retrosynthetic analysis .
69